BZOJ1736 [Usaco2005 jan]The Wedding Juicer 婚宴的榨汁机

从外面一点一点往里面拓展(floodfill)，每次找出最小的一个点，计算它对答案的贡献就好了。。。

找最小的点的话，直接pq就行

 /**************************************************************
     Problem: 1736
     User: rausen
     Language: C++
     Result: Accepted
     Time:196 ms
     Memory:2116 kb
 ****************************************************************/

 #include <cstdio>
 #include <queue>

 using namespace std;
 typedef long long ll;
 const int N = ;
 const int dx[] = {, , , -};
 const int dy[] = {, -, , };

 inline int read();

 struct data {
     int x, y, h;
     data(int _x = , int _y = , int _h = ) : x(_x), y(_y), h(_h) {}

     inline bool operator < (const data &d) const {
         return h > d.h;
     }
 };

 int n, m;
 int mp[N][N], v[N][N];
 priority_queue <data> h;

 inline ll work(){
     ll res = ;
     int x, y, k;
     data now;
     while (!h.empty()) {
         now = h.top(), h.pop();
         for (k = ; k < ; ++k) {
             x = now.x + dx[k], y = now.y + dy[k];
             if (x <=  || y <=  || x > n || y > m || v[x][y]) continue;
             v[x][y] = ;
             if (mp[x][y] < now.h)
                 res += now.h - mp[x][y], mp[x][y] = now.h;
             h.push(data(x, y, mp[x][y]));
         }
     }
     return res;
 }

 int main() {
     int i, j;
     m = read(), n = read();
     for (i = ; i <= n; ++i)
         for (j = ; j <= m; ++j) mp[i][j] = read();
     for (i = ; i <= n; ++i)
         for (j = ; j <= m; ++j)
             if (i ==  || j ==  || i == n || j == m)
                 h.push(data(i, j, mp[i][j])), v[i][j] = ;
     printf("%lld\n", work());
     return ;
 }

 inline int read() {
     static int x;
     static char ch;
     x = , ch = getchar();
     while (ch < '' || '' < ch)
         ch = getchar();
     while ('' <= ch && ch <= '') {
         x = x *  + ch - '';
         ch = getchar();
     }
     return x;
 }